College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 229: 33


(a) The cart is moving with a speed of $14.3~m/s$ at point 3. (b) The cart will reach point 4.

Work Step by Step

(a) We can use the conservation of energy to find the speed of the cart at point 3. WE can let $U_1 = 0$: $KE_3+U_3 = KE_1+U_1$ $KE_3 = KE_1+0-U_3$ $\frac{1}{2}mv_3^2 = \frac{1}{2}mv_1^2-mgh_3$ $v_3^2 = v_1^2-2gh_3$ $v_3 = \sqrt{v_1^2-2gh_3}$ $v_3 = \sqrt{(20.0~m/s)^2-(2)(9.81~m/s^2)(10.0~m)}$ $v_3 = 14.3~m/s$ The cart is moving with a speed of $14.3~m/s$ at point 3. (b) Let's assume that all the initial kinetic energy is converted to gravitational potential energy. We can find the maximum height the cart can reach: $U_g = KE_1$ $mgh = \frac{1}{2}mv^2$ $h = \frac{v^2}{2g}$ $h = \frac{(20.0~m/s)^2}{(2)(9.81~m/s^2)}$ $h = 20.4~m$ Since the cart can reach a maximum height of 20.4 meters, the cart will reach point 4 which is at a height of 20.0 meters.
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